Question

By method of mathematical induction, the value of $1^3+3^3+5^3\cdots n\text{terms}=n^2(x{n}^2-1)$ is true for $n\in N.$
Then $x$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let $P\left(n\right)=1^3+3^3+5^3\cdots n\text{terms}\forall n\in N$
so, it is also true for $n=1$
$P\left(1\right)=1$
From $RHS,$ we have $P\left(1\right)=1.(x-1)=1$
Hence on comparing we have $x=2$
Now, let $P\left(n\right):1^3+3^3+5^3\cdots n\text{terms}\forall n\in N$ is true
Check at $n=n+1$
$P(n+1):1^3+3^3+5^3\cdots +(2n-1{)}^3+(2n+1{)}^3=n^2(2n^2-1)+(2n+1{)}^3$
on expanding we get,
$P(n+1)=2n^4+8n^3+11n^2+6n+1=(n+1{)}^2[2(n+1{)}^2-1]$
So, $P(n+1)$ is true.